# Full text of "Mathematical And Physical Papers - Iii"

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```252     ON  THE  COMPOSITION  AND  RESOLUTION  OF  STREAMS OF
16.    It follows from this theorem that no partial analysis of light, such, for example, as would be produced by reflexion from the surface of glass or metal, or by transmission through a doubly absorbing medium, can from equivalent groups produce groups which   are  not  equivalent  to  each  other;   and  we   have  seen already that this cannot be done by means of the alteration of phase accompanying double refraction.    It follows, therefore, that equivalent groups are optically undistinguishable.
In proving this property of equivalent groups, it has been supposed that the polarizations of the two streams into which any group was resolved were opposite, such being the case in nature. But did a medium exist such that the two streams of light which it transmitted independently were polarized otherwise than oppositely, it would still not enable us to distinguish between equivalent groups.
17.    The experimental definition of common light is, light which is  incapable of exhibiting  rings of any kind when examined by a crystal of Iceland spar and an analyzer, or by some equivalent combination.    Consequently, a group of independent polarized streams will together be equivalent to common light when,  on  being resolved  in  any  manner  into  two  oppositely polarized pencils, the intensities of the two are the same, and of course equal to half that of the original group.    Accordingly,
•'                      in order that the group should be equivalent to common light,
it is necessary and sufficient that the constants B> 0, D should vanish.
18.    Let  us  now see  under what  circumstances   two  independent streams of polarized light can together be  equivalent to common light.
Let a, /3 refer to the first, and a', /3' to the second stream, and let the intensities of the two streams be as 1 to n; then we get from the formulae (16)
sin 2/3 -f n sin 2/3' = 0; cos 2a cos 2/3 -j- n cos 2a' cos 2/3' = 0 ; sin 2a cos 2/3 4- n sin 2a! cos 2/3' = 0.
Transposing, squaring, and adding, we find n* — l, and therefore ?i=l, since n is essentially positive. Since j3 and /3' are```